Math: A counting problem

Tutoring math 12 and some university courses, you get asked about counting. The tutor opens the discussion with a couple of examples.

Questions that ask, “How many ways can people be lined up for a photograph” or “how many five-card hands have two aces” or “how many ways can you fill out a multiple choice test” are all counting problems.

When an item, once used, cannot be used again, we call that without replacement. Counting without replacement can often be done using permute or choose, but there are other options.  For a quick summary of permute and choose, see my article here.

When an item can be used over again, we call that with replacement. A question with replacement is often done using an exponential expression, as we shall see.

Example 1: How many ways can you arrange the letters of the word radio?

Solution: The answer is 5P5, or 120. You use P – which means Permute – because the order of the letters is what matters, and because the process is without replacement: that is, in a given arrangement, each letter can only appear once.

Example 2: How many five card hands from a standard deck of 52 cards have exactly 2 aces?

Solution: Here, the order doesn’t matter, but the process is still without replacement: if you get the queen of hearts, you cannot draw it again in the same hand. Without replacement, and where order doesn’t matter, points to Choose (aka Combination): Specifically, (4C2)(48C3) or 103776. The reasoning is as follows: from the 4 aces, choose 2: 4C2. Then, from the other 48 non-ace cards, choose 3: 48C3. Multiply the two results together since they both happen yet are independent of each other. Independent means that the choosing of the two aces has no influence on which non-ace cards will be chosen.

Example 3: How many possible ways can a 20-question multiple choice test be filled out, if each question has five options?

Solution: Order matters here, but it is with replacement: for instance, you can choose answer A over and over again. The number of ways you can fill out the test is 5^{20}. The reasoning is that you can fill out the first question five possible ways, then the second five possible ways, and so on. Your possibilities are (5)(5)(5)(5)......(5)=5^{20}. The number is absurdly large, so we’ll just refer to it as 5^{20}.

I’ll say more about counting in future posts. Hope your exams are going well:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: Impulse

Tutoring physics, you get asked about impulse now and again.  The tutor introduces the topic, explaining its connection with momentum.

Impulse has many definitions.  Fundamentally, it means change in momentum.  Momentum is a vector quantity, which means it has a direction as well as a magnitude.  Impulse, being the change in momentum, is a vector as well.

A formula definition for impulse is

(1)   \begin{equation*}I=Ft\end{equation*}

where I=impulse, F=force, and t=time.

Generally, impulse is thought to mean a large force being applied over a very short time duration. A club hitting a golf ball is such a situation.

Example: A golf ball of mass 80.0g is struck by a golf club with a force of 90.0N. The duration of contact is 120.ms. Find

a) the impulse;

b) the velocity of the golf ball after being struck.

Solution:

a) the impulse is given by I=Ft. The time is 120.ms, where ms means milliseconds. Therefore, the time is 0.120s.

(2)   \begin{equation*}I=(90.0)(0.120)=10.8kgm/s\end{equation*}

So the impulse is 10.8kgm/s. Note that the unit of impulse is the same as momentum: Newtons times seconds (Ns), which equates to kgm/s.

b) the velocity of the golf ball can be found using the momentum formula

(3)   \begin{equation*}p=mv\end{equation*}

Here, p (strangely, but universally) stands for momentum,m for mass, and v for velocity.

Assuming the golf ball is at rest before it was struck, the impulse applied to it becomes its momentum, since its momentum was zero before. Therefore, its momentum after being struck is 10.8kgm/s. From there, we can find the velocity using p=mv, as follows:

(4)   \begin{equation*}10.8=0.0800v\end{equation*}

Note that the mass, being 80.0g, is 0.0800kg. You need base units – i.e., kilograms, metres, and seconds – in the momentum and impulse formulas.

We arrive at

(5)   \begin{equation*}\frac{10.8}{0.0800}=135=v\end{equation*}

So, the golf ball’s velocity is 135m/s as it leaves. In real life, air friction quickly takes effect, giving the ball the graceful arc golfers know.

I’ve never been golfing myself:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Systems of equations: a word problem

Tutoring math, you revisit this type of problem often.  The math tutor gives a quick reminder just ahead of exams.

Here’s the problem:

A saleswoman gets paid a base monthly salary plus commission.  One month she sells $50 000 and gets paid a total of $4200.  Another month she sells $75 000 and gets a total of $5700.  Find her base salary and commission rate.

Like any word problem, we begin with let statements:

(1)   \begin{equation*}\mbox{let x}=\mbox{ base salary}\end{equation*}

(2)   \begin{equation*}\mbox{let y}=\mbox{ commission rate}\end{equation*}

We set up the equations that reflect the two months’ earnings:

(3)   \begin{equation*}x+50 000y=4200\end{equation*}

(4)   \begin{equation*}x+75 000y=5700\end{equation*}

In a situation like this, substitution is easy. Looking at the first equation above, we notice that if x+50 000y=4200, then x=4200 - 50 000y. We can then substitute 4200 - 50 000y for x in the other equation:

(5)   \begin{equation*}(4200 - 50 000y) + 75 000y = 5700\end{equation*}

Continuing, we simplify the left side:

(6)   \begin{equation*}4200 - 50000y + 75000y=5700\end{equation*}

(7)   \begin{equation*}4200+25000y=5700\end{equation*}

Now we subtract 4200 from both sides:

(8)   \begin{equation*}25000y=1500\end{equation*}

Finally we divide both sides by 25000:

(9)   \begin{equation*}y=\frac{1500}{25000}=0.06\end{equation*}

Substitute the known value of y into either equation to find x:

(10)   \begin{equation*}x+50000(0.06)=4200\end{equation*}

Simplify:

(11)   \begin{equation*}x+3000=4200\end{equation*}

Subtract 3000 from both sides:

(12)   \begin{equation*}x=1200\end{equation*}

To answer the problem: the saleswoman’s base monthly salary is $1200, while her commission rate is 0.06. Note, of course, that 0.06 is the same as 6 percent.

Hope your exam prep is going well:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: a little more on factoring

Tutoring high school math, you might be in review mode now.  The math tutor brings up a couple of factoring instances.

Instance 1:  Suppose you need to factor -2x^2 + 4x + 70.  How would you proceed?

Solution: First, you would remove the common factor (for more explanation, see my post here). However, you are best to factor out the negative 2, rather than just the 2. In general, when factoring an expression with the lead term negative, you should factor out the negative as a common factor.

We proceed:

-2x^2+4x+70=-2(x^2-2x-35)

Next, we factor the inside expression using the easy trinomial method.

-2(x^2-2x-35)=-2(x-7)(x+5)

Sometimes you encounter an expression with just a negative in front.

Instance 2: Factor -x^2 +10x – 24

Solution: First, factor out the negative:

-x^2+10x-24=-(x^2-10x+24)

Now, factor inside, using the easy trinomial technique:

-(x^2-10x+24)=-(x-6)(x-4)

I’ll be mentioning more little tips and tricks the next few posts, with the hope of refreshing the skills towards first semester exams:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Physics: back emf

Tutoring physics 12, you encounter this concept.  The tutor offers the formula with an explanation.

Induction refers to the creation of voltage in a wire that is not connected to a power source.  It’s the premise that enables generation of electricity.  To induce voltage in a loop of wire, you need to spin it in a magnetic field.  The axis of rotation must be at right angles to the field.  You’ll get induced voltage according to

(1)   \begin{equation*}Emf=-N\frac{\delta(BA)}{\delta(t)}\end{equation*}

In the equation above, high school students may not recognize the \delta; you can substitute the more familiar Δ.

I’ll be covering the above formula in more detail in future posts, but today the focus is back emf.  Back emf evolves during the operation of an electric motor due to the fact that, once again, a loop of wire is spinning in a magnetic field.  Back emf is so-called because it opposes the line voltage powering the motor.  However, back emf is good, not bad.

Consider the following question:

The resistance of an electric motor is 10.0Ω, while the line voltage is 120V.  At operating speed, the motor draws 2.5A.  What is the back emf?

Solution:

We know that, without considering back emf, I=\frac{V}{R}.  Therefore, the current drawn should be I=\frac{120}{10.0}=12A.  The fact that, at operating speed, the motor draws much less than 12A, is due to back emf.  Specifically,

(2)   \begin{equation*}\mbox{Back emf= }V_{back}=V_{line}-IR\end{equation*}

In our specific case,

(3)   \begin{equation*}V_{back}=120 - (2.5)(10.0)=95V\end{equation*}

Essentially, as the coils of wire in the motor start to spin in the motor’s magnetic field, they become a generator. The generator opposes the line voltage. In our example, with V_{back}=95V, the line voltage is reduced from 120V to 25V in the motor. However, that’s good, because it drastically reduces the motor’s power consumption. Recalling the equation for power, P=I^2R, we see that without back emf, our motor would consume (12)^2(10.0) or 1440 Watts. Because of back emf, it consumes only (2.5)^2(10.0)=62.5W.

The efficiency of an electric motor can be related as

(4)   \begin{equation*}eff=\frac{V_{back}}{V_{line}}\end{equation*}

By that relationship, our motor above is \frac{95}{120} or 79% efficient.

In future posts I’ll be revisiting the concepts touched upon herein:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: factoring difference of cubes

The math tutor continues with factoring cubes.  Tutoring calculus, this formula is another standby.

Let’s turn to factoring x^3 – 64. Realizing that 64=4^3, you actually are facing x^3-4^3

You need the formula

a^3 – b^3=(a-b)(a^2 + ab + b^2)

For x^3-64, we substitute x for a and 4 for b. We arrive at

x^3-64=(x-4)(x^2+x(4)+16)=(x-4)(x^2+4x+16)

What about factoring 343x^6-8y^3? Since 343=7^3 and 8=2^3, we can rewrite the expression as
(7x^2)^3-(2y)^3

Carefully substituting 7x^2 for a and 2y for b, we arrive at

343x^6-8y^3=(7x^2-2y)((7x^2)^2+(7x^2)(2y)+(2y)^2))

Finally we simplify to

(7x^2-2y)(49x^4+14x^2y+4y^2)

The key with using these formulas is knowing how to rewrite your own expression so the proper substitution can be made.

Cheers,

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: factoring sum of cubes

Tutoring high school math, you rarely see this; it comes up often, however, in calculus.  The tutor offers a light treatment.

How do you factor x^3 + 125? You need the following formula:

a^3 + b^3 = (a + b)(a^2 – ab + b^2)

Of course, 125=5^3. Carefully substituting x for a and 5 for b, we arrive at

x^3 + 125=(x+5)(x^2-5x+25)

What about factoring 8x^3+1?

We need to realize that, really,

8x^3+1=(2x)^3 + 1^3

Substituting 2x for a and 1 for b in the formula, we arrive at

8x^3+1=(2x+1)((2x)^2 – 2x(1) + 1^2)=(2x+1)(4x^2 – 2x + 1)

For a hint about factoring the difference of cubes, please come back soon:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: How to handle multiple-stage fractions

Tutoring middle school or high school math, you come across these from time to time.  The math tutor explains how to unravel them.

We sometimes might encounter a fraction such as

(1)   \begin{equation*}\frac{\frac{3}{4}}{8}\end{equation*}

How do you simplify such a fraction to a normal, two-stage one?

First notice the longer of the two lines: it’s between the 34 and the 8. The long line can be instead written as a divided sign:

(2)   \begin{equation*}\frac{\frac{3}{4}}{8}=\frac{3}{4}\div{8}\end{equation*}

To divide with fractions, you rewrite the first, invert the second, and multiply. We proceed as follows:

(3)   \begin{equation*}(\frac{3}{4})(\frac{1}{8})=\frac{3}{32}\end{equation*}

Another example:

(4)   \begin{equation*}\mbox{Render }\frac{\frac{5}{9}}{\frac{4}{3}}\mbox{ to two stages }\end{equation*}

Although it’s subtle, you can see that the long line is between the 9 and the 4. Therefore, the fraction really means

(5)   \begin{equation*}\frac{5}{9}\div{\frac{4}{3}\end{equation*}

which really means

(6)   \begin{equation*}(\frac{5}{9})(\frac{3}{4})=\frac{15}{36}=\frac{5}{12}\end{equation*}

Have a great day:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Biology: the female system, cont.

When you tutor Biology 12, which is needed for nursing, you cover the male and female reproductive systems.  Here is the continuation from the Dec 30 post.

Previously I discussed the end of the monthly female cycle: if conception does not occur, menstruation does. However, what about the other case: when conception does occur?

The fertilized egg finds lodging in the blood lining of the uterus. Immediately, the placenta starts developing. As it does, it secretes a hormone (HCG) which orders the corpus luteum to continue producing progesterone. The progesterone from the corpus luteum maintains the blood lining in the uterus temporarily. Meanwhile, the placenta develops rapidly; soon, it starts producing its own progesterone. Throughout the pregnancy, the release of progesterone from the placenta will prevent menstruation.

There are a couple of terms that might be helpful. Implantation is the lodging of the fertilized egg in the blood lining of the uterus. The endometrium is the blood lining itself.

For an exploration of the hormonal influences at birthing, stay tuned:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: an assessment style problem

Tutoring math, you work with people heading back to school who need to pass assessment tests.  The math tutor demonstrates a problem they often ask about.

The problem is as follows:

Joe and Bill are scene painters for a theatre set.  Joe can paint the scene background by himself in 9 hours; Bill can do it by himself in 12.  How long will they take if they work together?

First, you realize that if Joe can paint the scene in 9 hours, he must work at the rate of 19 of the scene each hour. Similarly, if Bill can paint the scene in 12 hours, he must complete 112 of the scene per hour.

Add their hourly rates together to get their combined rate:

(1)   \begin{equation*}\frac{1}{9} + \frac{1}{12}=\frac{4}{36}+\frac{3}{36}=\frac{7}{36}\end{equation*}

Therefore, working together, they complete 736 of the job each hour. The time they will take to complete the one job is

(2)   \begin{equation*}1\ divided\ by\ \frac{7}{36}=(\frac{1}{1})(\frac{36}{7})=\frac{36}{7}=5\frac{1}{7}\ hours\end{equation*}

To check if the answer is reasonable, imagine the following: First, if Bill was as fast as Joe – i.e., they each could finish the scene, working solo, in 9 hours – then it makes sense that working together, they’d take half the time – i.e., 412 hours. Then again, if they both worked at Bill’s pace, either of them needing 12 hours to finish the job solo, then working together, they’d take half the time, or 6 hours. So if one works at the nine hour rate, the other at the twelve hour rate, and they work together, they take somewhere between 412 and 6 hours. Our answer of 517 sounds reasonable enough, from that point of view.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.